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Gravitation question

2004 · Q152
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Gravitation question

2004 · Q152

NEETPhysicsGravitationMCQ+4 / −1
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be
  1. A
    2R
  2. B
    4R
  3. C
    14{1 \over 4}41​R
  4. D
    12{1 \over 2}21​R
View written solutionFree

Correct answer: D

From equation of acceleration due to gravity.

ge=GMeRe2=G(4/3)πRe3Re2ρe{g_e} = {{G{M_e}} \over {R_e^2}} = {{G\left( {4/3} \right)\pi R_e^3} \over {R_e^2}}{\rho _e}ge​=Re2​GMe​​=Re2​G(4/3)πRe3​​ρe​

ge∝Reρe{g_e} \propto {R_e}{\rho _e}ge​∝Re​ρe​

Acceleration due to gravity of planet

gp∝Rpρp{g_p} \propto {R_p}{\rho _p}gp​∝Rp​ρp​

∴\therefore∴ Reρe=Rpρp⇒Reρe=Rp2ρe{R_e}{\rho _e} = {R_p}{\rho _p} \Rightarrow {R_e}{\rho _e} = {R_p}2{\rho _e}Re​ρe​=Rp​ρp​⇒Re​ρe​=Rp​2ρe​

⇒Re=12R \Rightarrow {R_e} = {1 \over 2}R⇒Re​=21​R   (∵\because∵ Re = R)

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